Solutions to Rudin Principles of Mathematical Analysis

134 pages

Solutions to Principles of Mathematical Analysis (Walter Rudin)

Jason Rosendale jason.rosendale@gmail.comDecember 19, 2011

This work was done as an undergraduate student: if you really don’t understand something in one of theseproofs, it is very possible that it doesn’t make sense because it’s wrong. Any questions or corrections can bedirected to jason.rosendale@gmail.com.

Exercise 1.1a

Let

r

be a nonzero rational number. We’re asked to show that

x

∈

Q

implies that (

r

+

x

)

∈

Q

. Proof of thecontrapositive:

→

r

+

x

is rational assumed

→

(

∃

p

∈

Q

)(

r

+

x

=

p

) deﬁnition of rational

→

(

∃

p,q

∈

Q

)(

q

+

x

=

p

) we’re told that

r

is rational

→

(

∃

p,q

∈

Q

)(

x

=

−

q

+

p

) existence of additive inverses in

Q

Because

p

and

q

are members of the closed additive group of

Q

, we know that their sum is amember of

Q

.

→

(

∃

u

∈

Q

)(

x

=

u

)

→

x

is rational deﬁnition of rationalBy assuming that

r

+

x

is rational, we prove that

x

must be rational. By contrapositive, then, if

x

is irrationalthen

r

+

x

is irrational, which is what we were asked to prove.

Exercise 1.1b

Let

r

be a nonzero rational number. We’re asked to show that

x

∈

Q

implies that

rx

∈

Q

. Proof of thecontrapositive:

→

rx

is rational assumed

→

(

∃

p

∈

Q

)(

rx

=

p

) deﬁnition of rational

→

(

∃

p

∈

Q

)(

x

=

r

−

1

p

) existence of multiplicative inverses in

Q

(Note that we can assume that

r

−

1

exists only because we are told that

r

is nonzero.) Because

r

−

1

and

p

are members of the closed multiplicative group of

Q

, we know that their product is also amember of

Q

.

→

(

∃

u

∈

Q

)(

x

=

u

)

→

x

is rational deﬁnition of rationalBy assuming that

rx

is rational, we prove that

x

must be rational. By contrapositive, then, if

x

is irrationalthen

rx

is irrational, which is what we were asked to prove.1

Exercise 1.2

Proof by contradiction. If

√

12 were rational, then we could write it as a reduced-form fraction in the form of

p/q

where

p

and

q

are nonzero integers with no common divisors.

→

pq

=

√

12 assumed

→

(

p

2

q

2

= 12)

→

(

p

2

= 12

q

2

)It’s clear that 3

|

12

q

2

, which means that 3

|

p

2

. By some theorem I can’t remember (possibly thedeﬁnition of ‘prime’ itself), if

a

is a prime number and

a

|

mn

, then

a

|

m

∨

a

|

n

. Therefore, since 3

|

pp

and 3 is prime,

→

3

|

p

→

9

|

p

2

→

(

∃

m

∈

N

)(

p

2

= 9

m

) deﬁnition of divisibility

→

(

∃

m

∈

N

)(12

q

2

= 9

m

) substitution from

p

2

= 12

q

2

→

(

∃

m

∈

N

)(4

q

2

= 3

m

) divide both sides by 3

→

(3

|

4

q

2

) deﬁnition of divisibilityFrom the same property of prime divisors that we used previously, we know that 3

|

4

∨

3

|

q

2

: itclearly doesn’t divide 4, so it must be the case that 3

|

q

2

. But if 3

|

qq

, then 3

|

q

∨

3

|

q

. Therefore:

→

(3

|

q

)And this establishes a contradiction. We began by assuming that

p

and

q

had no common divisors, but wehave shown that 3

|

p

and 3

|

q

. So our assumption must be wrong: there is no reduced-form rational number suchthat

pq

=

√

12.

Exercise 1.3 a

If

x



= 0 and

xy

=

xz

, then

y

= 1

y

= (

x

−

1

x

)

y

=

x

−

1

(

xy

) =

x

−

1

(

xz

) = (

x

−

1

x

)

z

= 1

z

=

z

Exercise 1.3 b

If

x



= 0 and

xy

=

x

, then

y

= 1

y

= (

x

−

1

x

)

y

=

x

−

1

(

xy

) =

x

−

1

x

= 1

Exercise 1.3 c

If

x



= 0 and

xy

= 1, then

y

= 1

y

= (

x

−

1

x

)

y

=

x

−

1

(

xy

) =

x

−

1

1 =

x

−

1

= 1

/x

Exercise 1.3 d

If

x



= 0, then the fact that

x

−

1

x

=1 means that

x

is the inverse of

x

−

1

: that is,

x

= (

x

−

1

)

−

1

= 1

/

(1

/x

).

Exercise 1.4

We are told that

E

is nonempty, so there exists some

e

∈

E

. By the deﬁnition of lower bound, (

∀

x

∈

E

)(

α

≤

x

):so

α

≤

e

. By the deﬁnition of upper bound, (

∀

x

∈

E

)(

x

≤

β

): so

e

≤

β

. Together, these two inequalities tell usthat

α

≤

e

≤

β

. We’re told that

S

is ordered, so by the transitivity of order relations this implies

α

≤

β

.2

Exercise 1.5

We’re told that

A

is bounded below. The ﬁeld of real numbers has the greatest lower bound property, so we’reguaranteed to have a greatest lower bound for

A

. Let

β

be this greatest lower bound. To prove that

−

β

isthe least upper bound of

−

A

, we must ﬁrst show that it’s an upper bound. Let

−

x

be an arbitrary element in

−

A

:

→ −

x

∈ −

A

assumed

→

x

∈

A

deﬁnition of membership in

−

A

→

β

≤

x β

= inf(

A

)

→ −

β

≥ −

x

consequence of 1.18(a)We began with an arbitrary choice of

−

x

, so this proves that (

∀−

x

∈ −

A

)(

−

β

≥ −

x

), which by deﬁnitiontells us that

−

β

is an upper bound for

−

A

. To show that

−

β

is the least such upper bound for

−

A

, we choosesome arbitrary element less than

−

β

:

→

α <

−

β

assumed

→ −

α > β

consequence of 1.18(a)Remember that

β

is the greatest lower bound of

A

. If

−

α

is larger than inf(

A

), there must be someelement of

A

that is smaller than

−

α

.

→

(

∃

x

∈

A

)(

x <

−

α

) (see above)

→

(

∃−

x

∈ −

A

)(

−

x > α

) consequence of 1.18(a)

→

!(

∀−

x

∈ −

A

)(

−

x

≤

α

)

→

α

is not an upper bound of

−

A

deﬁnition of upper boundThis proves that any element less than

−

β

is not an upper bound of

−

A

. Together with the earlier proof that

−

β

is an upper bound of

−

A

, this proves that

−

β

is the least upper bound of

−

A

.

Exercise 1.6a

The diﬃcult part of this proof is deciding which, if any, of the familiar properties of exponents are consideredaxioms and which properties we need to prove. It seems impossible to make any progress on this proof unless wecan assume that (

b

m

)

n

=

b

mn

. On the other hand, it seems clear that we can’t simply assume that (

b

m

)

1

n

=

b

m/n

:this would make the proof trivial (and is essentially assuming what we’re trying to prove).As I understand this problem, we have deﬁned

x

n

in such a way that it is trivial to prove that (

x

a

)

b

=

x

ab

when

a

and

b

are integers. And we’ve declared in theorem 1.21 that,

by deﬁnition

, the symbol

x

1

n

is the elementsuch that (

x

n

)

1

n

=

x

. But we haven’t deﬁned exactly what it might mean to combine an integer power like

n

and some arbitrary inverse like 1

/r

. We are asked to prove that these two elements do, in fact, combine in theway we would expect them to: (

x

n

)

1

/r

=

x

n/r

.Unless otherwise noted, every step of the following proof is justiﬁed by theorem 1.21.3